A template and examples for upper level physics students taking quantum mechanics at St. Mary's College of Maryland. Includes examples of kets and column vectors.
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{depreequantum}
\title{Quantum Homework Template}
\author{Erin De Pree}
\date{Fall 2019}
\usepackage{natbib}
\usepackage{graphicx}
\begin{document}
\maketitle
\section*{Suppressing Section Numbering}
If you are going to label your sections by problem number, then it is a good idea to use \verb+\section*+ instead of \verb+\section+ so you can suppress the section numbering. This way you won't write ``3 Problem 2''!
Here, I'm going to work through problem 1.5 from our text. This will show you how to write kets and such. I've also included a section on matrices so you can see how to write them up.
Be sure to include the \verb+depreequantum.sty+ style file as it loads the necessary packages.
\section*{Problem 1.5}
A beam of spin-1/2 particles is prepared in the state
\[ \ket{\psi} = \tfrac{2}{\sqrt{13}} \ket{+} + i \tfrac{3}{\sqrt{13}} \ket{-} \]
\paragraph{Part a} What are the possible results of a measurement of the spin component $S_z$, and with probabilities would they occur?
You will either measure $S_z = +\hbar/2$ (spin up) or $S_z = - \hbar/2$ (spin down).
The probability of measuring $S_z = +\hbar/2$ is
\begin{align*}
\mathcal{P}_+ &= \abs{ \braket{+}{\psi} }^2
= \abs{ \bra{+} \qty( \frac{2}{\sqrt{13}} \ket{+} + i \frac{3}{\sqrt{13}} \ket{-} ) }^2 \\
&= \abs{ \frac{2}{\sqrt{13}} \braket{+} + i \frac{3}{\sqrt{13}} \braket{+}{-} }^2\\
&= \abs{ \frac{2}{\sqrt{13}} + 0 }^2
= \boxed{ \frac{4}{13} \simeq 31\% }
\end{align*}
While the probability of measuring $S_z = - \hbar/2$ is
\begin{align*}
\mathcal{P}_- &= \abs{ \braket{-}{\psi} }^2
= \abs{ \bra{-} \qty( \frac{2}{\sqrt{13}} \ket{+} + i \frac{3}{\sqrt{13}} \ket{-} ) }^2 \\
&= \abs{ \frac{2}{\sqrt{13}} \braket{-}{+} + i \frac{3}{\sqrt{13}} \braket{-} }^2 \\
&= \abs{ 0 + i \frac{3}{\sqrt{13}} }^2
= \qty( -i\frac{3}{\sqrt{13}} ) \qty( i \frac{3}{\sqrt{13}} ) \\
&= -i^2 \frac{9}{13}
= \boxed{ \frac{9}{13} \simeq 69\% }
\end{align*}
\paragraph{Part b} What are the possible results of a measurement of the spin component $S_x$, and with probabilities would they occur?
Measure $S_x = \pm \hbar/2$.
Probability of measuring $S_x = +\hbar/2$ is
\begin{align*}
\mathcal{P}_{+x} &= \abs{ {}_x\braket{+}{\psi} }^2
= \abs{ \qty( \frac{1}{\sqrt{2}} \bra{+} + \frac{1}{\sqrt{2}} \bra{-} ) \qty( \frac{2}{\sqrt{13}} \ket{+} + i \frac{3}{\sqrt{13}} \ket{-} ) }^2 \\
&= \abs{ \frac{1}{\sqrt{2}} \frac{2}{\sqrt{13}} + i \frac{1}{\sqrt{2}} \frac{3}{\sqrt{13}} }^2
= \abs{ \frac{2 + 3i}{\sqrt{26}} }^2 \\
&= \frac{1}{26} \qty( 2 - 3i ) \qty( 2+3i )
= \frac{ 4 - 6i + 6i - 9i^2}{26} \\
&= \frac{4 + 9}{26}
= \frac{13}{26}
= \boxed{ \frac{1}{2} = 50\% }
\end{align*}
Probability of measuring $S_x = -\hbar/2$ is
\begin{align*}
\mathcal{P}_{-x} &= \abs{ {}_x\braket{-}{\psi} }^2
= \abs{ \qty( \frac{1}{\sqrt{2}} \bra{+} -\frac{1}{\sqrt{2}} \bra{-} ) \qty( \frac{2}{\sqrt{13}} \ket{+} + i \frac{3}{\sqrt{13}} \ket{-} ) }^2 \\
&= \abs{ \frac{1}{\sqrt{2}} \frac{2}{\sqrt{13}} - i \frac{1}{\sqrt{2}} \frac{3}{\sqrt{13}} }^2
= \abs{ \frac{2 - 3i}{\sqrt{26}} }^2 \\
&= \frac{1}{26} \qty( 2 + 3i ) \qty( 2-3i )
= \frac{ 4 - 6i + 6i - 9i^2}{26} \\
&= \frac{4 + 9}{26}
= \frac{13}{26}
= \boxed{ \frac{1}{2} = 50\% }
\end{align*}
\paragraph{Part c} Histograms! See \cref{fig:HistrogramsProblem1.5} for the histograms. Also note the figure appears on the next page because \LaTeX{} is automatically placing it where it makes most sense (after it is referenced).
\begin{figure}
\begin{center}
\begin{tikzpicture}[>=latex]
\begin{scope}
\draw[->] (0,0) -- (6,0) node[right]{$S_z$};
\draw[->] (0,0) -- (0,5) node[right]{$\mathcal{P}$};
\foreach \x / \spin / \prob in {2/-/0.31, 4/+/0.69}
{ \draw[fill=Grey] (\x,0) ++(-0.5, 0) rectangle ++(1,4*\prob) ++(-0.5,0) node[above]{$\prob$};
\draw (\x, -0.25) node[below]{$\spin \hbar/2$} -- ++(0, 0.5);
}
\foreach \y / \prob in {1/0.25, 2/0.50, 3/0.75, 4/1}
{ \draw (-0.2, \y) node[left]{$\prob$} -- ++(0.4,0); }
\end{scope}
\begin{scope}[xshift=9cm]
\draw[->] (0,0) -- (6,0) node[right]{$S_x$};
\draw[->] (0,0) -- (0,5) node[right]{$\mathcal{P}$};
\foreach \x / \spin / \prob in {2/-/0.5, 4/+/0.5}
{ \draw[fill=Grey] (\x,0) ++(-0.5, 0) rectangle ++(1,4*\prob) ++(-0.5,0) node[above]{$\prob$};
\draw (\x, -0.25) node[below]{$\spin \hbar/2$} -- ++(0, 0.5);
}
\foreach \y / \prob in {1/0.25, 2/0.50, 3/0.75, 4/1}
{ \draw (-0.2, \y) node[left]{$\prob$} -- ++(0.4,0); }
\end{scope}
\end{tikzpicture}
\end{center}
\caption{Probabilities of measuring various spins in problem 1.5. You may find it easier to make your histograms in Excel or Google Sheets or by hand and include them as a picture.}
\label{fig:HistrogramsProblem1.5}
\end{figure}
\section*{Matrix Notation}
Going to use $\dot{=}$ to mean \emph{represented by}. In \LaTeX\,, use the notation: \verb+\dot{=}+. The quantum state isn't a column vector, but it can be represented by a column vector
\begin{align}
\ket{+} &\dot{=} \mqty( 1 \\ 0 )
& \ket{-} &\dot{=} \mqty( 0 \\ 1)
\label{eq:basismatrix}
\end{align}
This is the $z$-basis set.
Let's use the column vector notation to represent a state $\ket{\psi}$ as a projection of the state on the $\ket{+}$ and the projection on the $\ket{-}$:
\[ \ket{\psi} \dot{=} \mqty( \braket{+}{\psi} \\ \braket{-}{\psi} ) \]
\[ \ket{\psi} = a \ket{+} + b \ket{-} \]
\[ \ket{\psi} \dot{=} \mqty( a \\ b ) \]
\[ \bra{\psi} \dot{=} \mqty( a^\ast & b^\ast ) \]
We can also write the $x$ and $y$ kets in the $z$-basis:
\begin{align*}
\ket{+}_x &\dot{=} \frac{1}{\sqrt{2}} \mqty( 1 \\ 1)
& \ket{+}_y&\dot{=} \frac{1}{\sqrt{2}} \mqty( 1 \\ i ) \\
\ket{-}_x &\dot{=} \frac{1}{\sqrt{2}} \mqty( 1 \\ -1 )
& \ket{-}_y &\dot{=} \frac{1}{\sqrt{2}} \mqty( 1 \\ -i )
\end{align*}
\end{document}